Extending an Independent Set to Basis Theorem

Theorem

1. Suppose $V$ has a finite spanning set $T=\{y_1,\dots ,y_n\}$
2. Suppose $S=\{x_1,\dots ,x_k\}$ is a linearly independent subset of $V$
3. There is always a finite basis $S^{\prime }$ of $V$ such that $S\subset S^{\prime }$ The idea is set $S^{\prime }=S$

Proof

Process

1. If $S^{\prime }$ is a basis, then we’re done. Otherwise, union a vector of $T$ with the set of $S$ to create a new linear independent set $S^{\prime }=S\cup \{{\vec{y}}_i\}$/ (Read Linear Independence)
2. Repeat until $S^{\prime }$ is a basis

Proof

1. Assume $S^{\prime }=S$ is not a basis
2. We know $S^{\prime }$ is Linear Independence therefore, $span(S^{\prime })=V$
3. We find, $y_i\in T$ s.t ${\vec{y}}_i\notin S^{\prime }$ (This is because if all of the guys in $S$ are in $T$, then $span(S)=V$ because $T$ is a spanning set)
4. It follows by the extension of Linear Independence Extension Lemma that $S^{\prime }\cup \{{\vec{y}}_2\}$ is linearly independent
5. Run the algorithm again. Now consider $S^{\prime }\cup \{{\vec{y}}_2\}=S$
6. The finiteness of T means that we run this algorithm finitely many times. Therefore, the final $S$ has at most $k+n$ elements which is finite.